Method X
On 8/3/23 02:54, Kevin Venzke wrote:
I plan to implement Method X when I get a moment. I'm not completely
skeptical about its monotonicity since it doesn't seem like there's an
obvious argument why it shouldn't be monotone.
Thanks :-)
In case it'll be of help, my implementation is here:
https://github.com/kristomu/quadelect/tree/master/src/singlewinner/dmt
strat_ifpp.h and .cc
I'll check the different Landau implementations to find out if mine is
busted.
-km
Method X is cloneproof as well! That makes it even more superior to Kemeny
(I think that cloneproofness is a must for a good electoral method). I also
think that IIB is met, because if a given ballot is not exhausted, it is
not irrelevant.
I'm building something similar to the quadelect program so I can check
similar things; indeed the results show that Smith//Method_X is monotonic
(if ties are resolved properly), cloneproof and DMTBR-compliant (and of
course ISDA-compliant). Finding a method that combines it all is an amazing
achievement! A polytime version, if such exists, would certainly be the
Holy Grail Method. If it doesn't exist, then the non-polytime version would
probably still be the best option - as it was already mentioned, real-life
elections wouldn't be at all as computation-costly as it is theoretically
possible.
Perhaps basing the elimination process on first. pref. Copeland score or
"friendly" score instead of simple plurality score should be considered - I
guess it might at least make Method X naturally ISDA-compliant, but the
criteria compliance and strategy resistance would need to be checked again.
Below I present the closest thing to a non-monotonicity example that I
found; by the way, you can check if the X score results are correct (and
the implementation is OK):
1: A>B>D>C
1: B>C>A>D
1: B>C>D>A
1: C>A>B>D
1: C>B>A>D
1: C>D>A>B
1: D>A>B>C
2: D>C>A>B
1: D>C>B>A
The scores are A:0, B:5, C:5, D:5, so there's a B-C-D tie.
But then we raise B by changing CBAD to BCAD and DCBA to DBCA.
The new scores are A:4, B:5, C:0, D:6, so D wins - B is no longer in the
winning tie.
So if we want to 100% guarantee the monotonicity in all edge cases, we
should make sure that such a tie is not resolved by making B the winner.
śr., 2 sie 2023 o 01:21 Kristofer Munsterhjelm km_elmet@t-online.de
napisał(a):
This is method X, the monotone burial-resistant method from the previous
post: (It doesn't really have a name yet.)
-
Each candidate A obtains his score by eliminating candidates in
rounds, one candidate per round, until one other candidate (B) remains
or A himself is forced to be eliminated. In the former case, A's score
is A>B. In the latter case, A's score is zero. -
When figuring out A's score, the method chooses the sequence of
candidates to eliminate so as to maximize that score. -
In a round, the method can never eliminate a candidate who has more
than 1/n of the total number of (non-exhausted) first preferences, where
n is the number of remaining candidates in the round in question. -
Unlike IFPP, it never eliminates more than one candidate per round.
-
Highest score wins.
That's it!
Now for the bad news:
- it's not summable (a summary takes O(n2^n) space),
- it's not polytime (ditto),
- its use of quotas means it would probably fail IIB,
- and I have no idea why it works.
I can't say I wasn't tempted to name it after myself since it's kind of
a big deal (if I can verify its monotonicity), but given its drawbacks,
maybe it's not a good idea? It's like the Kemeny of monotone
burial-resistant methods: slow and impractical, but it shows what's
possible.
I would very much like some independent verification or replication,
though.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
On 8/3/23 20:17, Filip Ejlak wrote:
Method X is cloneproof as well! That makes it even more superior to
Kemeny (I think that cloneproofness is a must for a good electoral
method). I also think that IIB is met, because if a given ballot is
not exhausted, it is not irrelevant.
I'm building something similar to the quadelect program so I can check
similar things; indeed the results show that Smith//Method_X is
monotonic (if ties are resolved properly), cloneproof and
DMTBR-compliant (and of course ISDA-compliant). Finding a method that
combines it all is an amazing achievement! A polytime version, if such
exists, would certainly be the Holy Grail Method. If it doesn't exist,
then the non-polytime version would probably still be the best option -
as it was already mentioned, real-life elections wouldn't be at all as
computation-costly as it is theoretically possible.
Nice! It's always a good thing to get more evidence that it works!
Perhaps basing the elimination process on first. pref. Copeland score or
"friendly" score instead of simple plurality score should be considered
- I guess it might at least make Method X naturally ISDA-compliant, but
the criteria compliance and strategy resistance would need to be checked
again.
Another idea I've been bouncing around in my mind is to do it Tideman's
Alternative style, i.e. if there are candidates in the current round who
aren't in the Smith set (based on the remaining candidates), then
eliminate all of them before continuing the procedure. It would be more
strict than Smith//Method X, but it would also eliminate more candidates
at once, easing the computational burden. Like your proposals, I don't
know if it would be monotone or keep the strategy resistance.
Below I present the closest thing to a non-monotonicity example that I
found; by the way, you can check if the X score results are correct (and
the implementation is OK):
1: A>B>D>C
1: B>C>A>D
1: B>C>D>A
1: C>A>B>D
1: C>B>A>D
1: C>D>A>B
1: D>A>B>C
2: D>C>A>B
1: D>C>B>A
The scores are A:0, B:5, C:5, D:5, so there's a B-C-D tie.
But then we raise B by changing CBAD to BCAD and DCBA to DBCA.
The new scores are A:4, B:5, C:0, D:6, so D wins - B is no longer in the
winning tie.
So if we want to 100% guarantee the monotonicity in all edge cases, we
should make sure that such a tie is not resolved by making B the winner.
I can verify that the scores are 0, 5, 5, 5 before, and that after
raising B they're 4, 5, 0, 6.
I briefly experimented with using first preferences as a tiebreaker; it
seemed to work -- I didn't get any pushover strategies, and indeed
Plurality's order for the example above is D>C>B>A, keeping B from first
place. However, it did slightly increase strategic susceptibility, and
if you're doing things like Smith//, it could add a remote chance of
nonmonotonicity through usual Smith//Plurality shenanigans.
I'm thinking that a more consistent or elegant tiebreaker is to use
leximax: find the next highest score obtainable, then the third highest,
etc. But it's a real pain to implement (particularly to pass through
vectors of scores recursively), so I haven't tried yet.
I can't prove that either is monotone because I don't know why method X
itself is monotone yet :-)
-km
On 8/3/23 21:20, Kristofer Munsterhjelm wrote:
I'm thinking that a more consistent or elegant tiebreaker is to use
leximax: find the next highest score obtainable, then the third highest,
etc. But it's a real pain to implement (particularly to pass through
vectors of scores recursively), so I haven't tried yet.
That might be much harder to do than I thought because you would want to
avoid the Kemeny-esque clone problems that could result - e.g. adding a
bunch of clones could give one candidate more high scores than another
candidate due to there being more ways to eliminate the clones than before.
So perhaps it's better to consider the first preference scores
throughout the optimal elimination order, e.g.
if the elimination order is "eliminate D, then C, then score is A>B"
then the extended score vector could be
[A>B, (first preference of A when D but not C is eliminated),
A's first preferences]
though that could still produce clone problems, possibly... it's hard to
reason about what would.
A fourth option is to relax the quota, e.g. make it 1/3 instead of 1/n
and see which candidate gets the highest score, and use that as a
tiebreaker. But it feels ad hoc. Quotas higher than 1/n still work,
but used on their own (i.e. as methods by themselves, not as
tiebreakers), the method X analogs are much more susceptible to strategy.
-km
Assets in the general sense that potentially transferable first place votes
can be considered as assets as opposed to liabilities.... and proxy in the
sense of easing the strategic burden on the voters via (for example) voting
for a public ranking ... it seems like creative use of DVV in conjunction
with these other burden reducing features might side step the main problem
of plain asset voting... in the spirit of brainstorming inspired by your
creative Method X.
On Wed, Aug 2, 2023, 1:54 PM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
Responding to both DSV posts in one message...
On 8/2/23 14:46, Forest Simmons wrote:
On Tue, Aug 1, 2023, 4:21 PM Kristofer Munsterhjelm
<km_elmet@t-online.de mailto:km_elmet@t-online.de> wrote:
This is method X, the monotone burial-resistant method from the
previous
post: (It doesn't really have a name yet.)
- Each candidate A obtains his score by eliminating candidates in
rounds, one candidate per round, until one other candidate (B)
remains
or A himself is forced to be eliminated. In the former case, A's
score
is A>B. In the latter case, A's score is zero.
- When figuring out A's score, the method chooses the sequence of
candidates to eliminate so as to maximize that score.
This DSV feature could be the key to overcoming our main objection to
Asset Voting ... the loose cannon proxy.
I'm not entirely sure how, could you explain? I can maximize for A
because I can infer what A's trying to maximize (his score) and the
means by which it can be done (choosing a particular sequence of
eliminations).
But in Asset Voting (and delegable proxy) it would seem like the main
benefit is that you can't specify your preferences in exact detail, so
you instead choose to give your voting power to someone who can.
Then the loose cannon is a problem where someone you trusted goes and
does the wrong thing. Because the system doesn't know your preferences
(if it did, you wouldn't need to use a proxy), it can't preemptively
optimize around the "glitches".
Or am I thinking of something entirely different?
Restricting to Smith requires lots of first place vote transfers ...
right?
Yeah, that's right; that's why it's so surprising that Smith//X would
stay monotone. Dropping a bunch of non-Smith candidates changes the lay
of the land in unpredictable ways -- yet apparently not so unpredictable
that monotonicity is affected. "Where is the deeper pattern in this chaos?"
So what about (for each candidate X) just bequeathing all of X's
first place assets to the Smith member Y against which X has the
fewest losing votes?
You mean for making a simpler method that still passes Smith and ISDA?
Smith//X would automatically remove all non-Smith members before doing
X, and method X itself finds a maximally beneficial elimination order
for each candidate, thus doing something disadvantageous to candidate X
wouldn't seem to have a place in that procedure.
-km
On 8/3/23 02:54, Kevin Venzke wrote:
I plan to implement Method X when I get a moment. I'm not completely
skeptical about its monotonicity since it doesn't seem like there's an
obvious argument why it shouldn't be monotone.
For Landau, I use some code like this, which I base on the explanation on Wikipedia:
let landau = [];
for(x in candidates) {
let i_am_covered = false;
for(z in candidates) {
if( x != z ) { // X can't cover themselves
let found_y = false;
for(y in candidates) {
// Note that beats_or_ties should evaluate as true if the two operands are the same candidate
if( x beats_or_ties y && y beats_or_ties z ) {
// alternative to the above line which should be right:
// if( not( y beats x || z beats y ) ) {
found_y = true; // found a counterexample to Z covering X
break; // cease checking for Ys for this Z
} //if x...
} //for y
if( not found_y ) {
i_am_covered = true; // Z evidently covers X
break; // cease checking for Zs for this X
} // if not found_y
} // if x!=z
} // for z
if( not i_am_covered ) landau.append( x );
} // for x
return landau;
I've run my Landau calculator against this for a day, and I haven't
found any discrepancies. So it looks like my Landau calculator does
work, and it's good to know that the weird things I was seeing weren't
due to bugs.
My Landau set function uses short beatpaths; I was concerned that the
beatpath calculation code itself was wrong.
In any event, here's the data for Landau//Method X. It's clearly not
monotone, but as I said elsewhere, I'm not surprised since Forest proved
that was inevitable as (under reasonable assumptions) independence of
covered alternatives implies nonmonotonicity. But it's not strategy
resistant either, so there's not much of a reason to use it!
Burial, no compromise: 9740 0.199174
Compromise, no burial: 7624 0.155904
Burial and compromise: 3730 0.076275
Two-sided: 5483 0.112122
Other coalition strats: 22322 0.456464
---=========
Manipulable elections: 48899 0.999939
Landau,X isn't much better:
Burial, no compromise: 6093 0.124744
Compromise, no burial: 10123 0.207252
Burial and compromise: 1277 0.0261445
Two-sided: 7138 0.146139
Other coalition strats: 23721 0.485648
---=========
Manipulable elections: 48352 0.989927
So that's not the way to get uncovered.
-km
Hi Kristofer,
It wasn't so easy, but regrettably I think I have a monotonicity counter-example:
408: B>C>A
329: A>C>B
126: C>A>B
91: C>B>A
43: A>B>C --> B>A>C
(total 997)
For the first round, A and B votes both exceed 1/3rd (332.33) and so only C can be
eliminated.
The match-up A:B gives B a very slight win of 499 vs 498 for A. C can't score anything.
Scores: B 499, A 498, C 0.
Now change the 43 to B>A>C, theoretically helping B further.
First round totals become 329 A, 451 B, 217 C. So it is now allowed to eliminate A.
Both A and B fare worse against C than against each other and so prefer to score off of
eliminating C.
B improves its score to 542 while A's score is reduced to 455.
However, when A is eliminated, C can score 546 from their matchup with B.
New scores: C 546, B 542, A 455.
Let me know if you think I have the method wrong.
One thing I noticed is that modifying the quota rule allows you at one extreme to
implement IRV (i.e. by saying that only the candidate with the fewest votes can be
eliminated each round) and at the other extreme to implement "max votes-for wins" (by
imposing no quota requirement at all). While the latter is monotone, it doesn't satisfy
majority favorite. It's interesting to consider whether any quota rule could at least
preserve monotonicity and add majority favorite. I'm thinking no, though.
Kevin
votingmethods.net
On 8/5/23 05:49, Kevin Venzke wrote:
Hi Kristofer,
It wasn't so easy, but regrettably I think I have a monotonicity counter-example:
408: B>C>A
329: A>C>B
126: C>A>B
91: C>B>A
43: A>B>C --> B>A>C
(total 997)
For the first round, A and B votes both exceed 1/3rd (332.33) and so only C can be
eliminated.
The match-up A:B gives B a very slight win of 499 vs 498 for A. C can't score anything.
Scores: B 499, A 498, C 0.
I can verify that the scores are B: 499 > A: 498 > C: 0.
Now change the 43 to B>A>C, theoretically helping B further.
First round totals become 329 A, 451 B, 217 C. So it is now allowed to eliminate A.
Both A and B fare worse against C than against each other and so prefer to score off of
eliminating C.
B improves its score to 542 while A's score is reduced to 455.
However, when A is eliminated, C can score 546 from their matchup with B.
New scores: C 546, B 542, A 455.
And I can verify that the scores are C: 546 > B: 542 > A: 455.
Well done. Well, I would rather have wanted it to be monotone, but it's
better to know the truth! I guess that makes this "very low
nonmonotonicity" rather than monotone - now I know how the IRVists feel
when people complain about nonmonotonicity!
Here's a minimal example produced by linear programming:
1: A>B>C
7: A>C>B
8: B>A>C
3: C>A>B
4: C>B>A
the scores are B: 12 > A: 11 > C: 0, then after changing ABC to BAC the
scores become C: 14 > B: 13 > A: 10.
Interestingly, for your example, fpA-fpC says that the correct ordering
for the "before" election is C>B>A, whle Carey says B>A>C. My example,
on the other hand, doesn't have this distinction... but it has a
Condorcet cycle both before and after, thus showing that Smith//X won't
solve the problem.
Despite the example showing that X itself isn't monotone, I'm more
confident now that (properly phrased) DMTBR is compatible with both
monotonicity and Condorcet. Prior to method X, we only had the fpA-fpC
generalizations, IFPP, and IRV; the first were only DMTCBR, while the
latter two were clearly nonmonotone. I was worried that there might be
an impossibility theorem of some kind proving that monotonicity would be
forever out of our grasp for burial-resistant Condorcet methods.
I can also use method X to find out just what kind of DMTBR should hold,
and then build off that. I think I have another idea that could work,
but it would be so incredibly ugly - basically "IRV with donations".
Or we could try to find out why X comes so close to monotonicity, since
it's the closest we've got so far. Doing so would require figuring out
why max A>B ("max votes-for") is monotone, and why Smith//method X also
seems to be (nearly) monotone, I think.
One thing I noticed is that modifying the quota rule allows you at one extreme to
implement IRV (i.e. by saying that only the candidate with the fewest votes can be
eliminated each round) and at the other extreme to implement "max votes-for wins" (by
imposing no quota requirement at all). While the latter is monotone, it doesn't satisfy
majority favorite.
That's right; making the quota more loose (i.e. giving the method more
candidates to choose eliminations from in a given round) doesn't seem to
hurt monotonicity until you go past 1/3, but it does hurt strategy
resistance. Going in the other direction is not strictly possible
because if you're in an n-way tie, every candidate has exactly 1/n of
the first preferences. So you would then need to also allow eliminating
the lowest scorer no matter what. This would make it more like IRV and
thus compromise its monotonicity (further).
It's interesting to consider whether any quota rule could at least
preserve monotonicity and add majority favorite. I'm thinking no, though.
The weakest quota I can think of that will preserve majority is 1/2.
Suppose A is voted first by a majority. Then A can never be eliminated,
so for any other candidate B, it eventually ends up being A vs B, and
since A is a majority favorite, A then wins. However, this is not
strategy resistant; even a constant quota of 1/3 for everything but the
final round (which is what I tried first) destroys strategy resistance.
Furthermore, as mentioned above, there seems to be a strange
relationship between the quota and the degree of nonmonotonicity - at
least if "Other" is a good indicator. For a three-candidate election,
1/3 is equivalent to "normal" method X, which we now know is (barely)
nonmonotone. However, loosening the quota to 1/2 introduces more
nonmonotonicity; then getting rid of the quota altogether gets us back
into the monotone domain.
E.g. with fixed quota 1/2, impartial culture, 5 candidates, 97 voters,
7500 elections:
Burial, no compromise: 218 0.0305793
Compromise, no burial: 1138 0.15963
Burial and compromise: 435 0.0610184
Two-sided: 5297 0.743021
Other coalition strats: 41 0.00575116
---=========
Manipulable elections: 7129 1
and with fixed quota 1/3:
Burial, no compromise: 403 0.0558017
Compromise, no burial: 1544 0.213791
Burial and compromise: 86 0.0119081
Two-sided: 5149 0.71296
Other coalition strats: 0 0
---=========
Manipulable elections: 7182 0.994461
-km
On Sat, Aug 5, 2023, 4:52 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
On 8/5/23 05:49, Kevin Venzke wrote:
Hi Kristofer,
It wasn't so easy, but regrettably I think I have a monotonicity
counter-example:
408: B>C>A
329: A>C>B
126: C>A>B
91: C>B>A
43: A>B>C --> B>A>C
(total 997)
For the first round, A and B votes both exceed 1/3rd (332.33) and so
only C can be
eliminated.
The match-up A:B gives B a very slight win of 499 vs 498 for A. C can't
score anything.
Scores: B 499, A 498, C 0.
I can verify that the scores are B: 499 > A: 498 > C: 0.
Now change the 43 to B>A>C, theoretically helping B further.
First round totals become 329 A, 451 B, 217 C. So it is now allowed to
eliminate A.
Both A and B fare worse against C than against each other and so prefer
to score off of
eliminating C.
B improves its score to 542 while A's score is reduced to 455.
However, when A is eliminated, C can score 546 from their matchup with B.
New scores: C 546, B 542, A 455.
And I can verify that the scores are C: 546 > B: 542 > A: 455.
Well done. Well, I would rather have wanted it to be monotone, but it's
better to know the truth! I guess that makes this "very low
nonmonotonicity" rather than monotone - now I know how the IRVists feel
when people complain about nonmonotonicity!
Here's a minimal example produced by linear programming:
1: A>B>C
7: A>C>B
8: B>A>C
3: C>A>B
4: C>B>A
the scores are B: 12 > A: 11 > C: 0, then after changing ABC to BAC the
scores become C: 14 > B: 13 > A: 10.
Interestingly, for your example, fpA-fpC says that the correct ordering
for the "before" election is C>B>A, whle Carey says B>A>C. My example,
on the other hand, doesn't have this distinction... but it has a
Condorcet cycle both before and after, thus showing that Smith//X won't
solve the problem.
Despite the example showing that X itself isn't monotone, I'm more
confident now that (properly phrased) DMTBR is compatible with both
monotonicity and Condorcet. Prior to method X, we only had the fpA-fpC
generalizations, IFPP, and IRV; the first were only DMTCBR, while the
latter two were clearly nonmonotone. I was worried that there might be
an impossibility theorem of some kind proving that monotonicity would be
forever out of our grasp for burial-resistant Condorcet methods.
Do you consider Implicit Approval Chain Climbing to be burial resistant?
It certainly punishes the burying faction on all of the examples Chris gave
highlighting his MinLosingVotes Pairwise Margins method.
As a reminder it is the only Banks efficient monotone, clone independent,
Universal Domain method that we know of so far.
In general, Agenda Based Chain Climbing is monotone when the agenda
formation is monotone ... so Borda and Kemeny Chain Climbing are also Banks
efficient monotone methods that are probably burial resistant, but neither
one is clone proof.
In general, elimination with "take down" is Banks efficient ... but not
monotone unless based on a fixed (no renormalization between eliminations)
monotone agenda.
Implicit Approval is monotone and clone proof and UD, but just barely UD.
It is maddenly frustrating trying to find another UD monotone, clone proof
agenda forming method.
I can also use method X to find out just what kind of DMTBR should hold,
and then build off that. I think I have another idea that could work,
but it would be so incredibly ugly - basically "IRV with donations".
Or we could try to find out why X comes so close to monotonicity, since
it's the closest we've got so far. Doing so would require figuring out
why max A>B ("max votes-for") is monotone, and why Smith//method X also
seems to be (nearly) monotone, I think.
One thing I noticed is that modifying the quota rule allows you at one
extreme to
implement IRV (i.e. by saying that only the candidate with the fewest
votes can be
eliminated each round) and at the other extreme to implement "max
votes-for wins" (by
imposing no quota requirement at all). While the latter is monotone, it
doesn't satisfy
majority favorite.
That's right; making the quota more loose (i.e. giving the method more
candidates to choose eliminations from in a given round) doesn't seem to
hurt monotonicity until you go past 1/3, but it does hurt strategy
resistance. Going in the other direction is not strictly possible
because if you're in an n-way tie, every candidate has exactly 1/n of
the first preferences. So you would then need to also allow eliminating
the lowest scorer no matter what. This would make it more like IRV and
thus compromise its monotonicity (further).
It's interesting to consider whether any quota rule could at least
preserve monotonicity and add majority favorite. I'm thinking no, though.
The weakest quota I can think of that will preserve majority is 1/2.
Suppose A is voted first by a majority. Then A can never be eliminated,
so for any other candidate B, it eventually ends up being A vs B, and
since A is a majority favorite, A then wins. However, this is not
strategy resistant; even a constant quota of 1/3 for everything but the
final round (which is what I tried first) destroys strategy resistance.
Furthermore, as mentioned above, there seems to be a strange
relationship between the quota and the degree of nonmonotonicity - at
least if "Other" is a good indicator. For a three-candidate election,
1/3 is equivalent to "normal" method X, which we now know is (barely)
nonmonotone. However, loosening the quota to 1/2 introduces more
nonmonotonicity; then getting rid of the quota altogether gets us back
into the monotone domain.
E.g. with fixed quota 1/2, impartial culture, 5 candidates, 97 voters,
7500 elections:
Burial, no compromise: 218 0.0305793
Compromise, no burial: 1138 0.15963
Burial and compromise: 435 0.0610184
Two-sided: 5297 0.743021
Other coalition strats: 41 0.00575116
---=========
Manipulable elections: 7129 1
and with fixed quota 1/3:
Burial, no compromise: 403 0.0558017
Compromise, no burial: 1544 0.213791
Burial and compromise: 86 0.0119081
Two-sided: 5149 0.71296
Other coalition strats: 0 0
---=========
Manipulable elections: 7182 0.994461
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
Hi Forest, is IACC the same thing as TACC (i.e. initial head is the least approved)? Because I don't find TACC to be monotone.
I actually don't know or haven't implemented any monotone Landau method. Not that I have really looked hard for one.
Kevin
(end)
Le samedi 5 août 2023 à 12:21:22 UTC−5, Forest Simmons forest.simmons21@gmail.com a écrit :
Do you consider Implicit Approval Chain Climbing to be burial resistant?
It certainly punishes the burying faction on all of the examples Chris gave highlighting his MinLosingVotes Pairwise Margins method.
As a reminder it is the only Banks efficient monotone, clone independent, Universal Domain method that we know of so far.
In general, Agenda Based Chain Climbing is monotone when the agenda formation is monotone ... so Borda and Kemeny Chain Climbing are also Banks efficient monotone methods that are probably burial resistant, but neither one is clone proof.
In general, elimination with "take down" is Banks efficient ... but not monotone unless based on a fixed (no renormalization between eliminations) monotone agenda.
Implicit Approval is monotone and clone proof and UD, but just barely UD. It is maddenly frustrating trying to find another UD monotone, clone proof agenda forming method.